Question

在我正在研究的金字塔应用程序中，我有以下情况：

class Widget(Base):
    __tablename__ = 'widgets'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))
    sidebar = Column(mysql.TINYINT(2))

    def __init__(self, name, sidebar):
        self.name = name
        self.sidebar = sidebar

class Dashboard(Base):
    __tablename__ = 'dashboard'
    user_id = Column(Integer, ForeignKey('users.id'), primary_key=True)
    widget_id = Column(Integer, ForeignKey('widgets.id'), primary_key=True)
    delta = Column(mysql.TINYINT)

    widget = relationship('Widget')

    def __init__(self, user_id, widget_id, delta):
        self.user_id = user_id
        self.widget_id = widget_id
        self.delta = delta 

class User(Base):
    __tablename__ = 'users'
    id = Column(Integer, primary_key=True)
    login = Column(Unicode(255), unique=True)
    password = Column(Unicode(60))
    fullname = Column(Unicode(100))

    dashboard = relationship('Dashboard', order_by='Dashboard.widget.sidebar, Dashboard.delta')

    def __init__(self, login, password, fullname):
        self.login = login
        self.password = crypt.encode(password)
        self.fullname = fullname

因此，我希望用户“仪表板”关系具有用户的仪表板记录，但是由“侧栏”（这是仪表板的关系属性）排序。目前我收到此错误：

sqlalchemy.exc.InvalidRequestError: Property 'widget' is not an instance of ColumnProperty (i.e. does not correspond directly to a Column).

关系声明中是否可以进行此排序？

谢谢！

Answer 1

有了这个，试着想一想SQL SQLAlchemy在尝试加载User.dashboard时应该发出什么。像SELECT * FROM dashboard JOIN widget ... ORDER BY widget.sidebar一样？还是SELECT * FROM dashboard ORDER BY (SELECT sidebar FROM widget...？通过不同的表对结果进行排序对relationship()的工作来说是开放式的，无法决定它自己。这样做的方法是提供一个Dashboard的列表达式，它可以提供这种顺序，当ORM对仪表板的表发出一个简单的SELECT时，以及当它在一个不是这样的时候引用它时-simple SELECT可以同时加入User，Dashboard表（例如，急切加载）。

我们提供自定义SQL表达式，特别是那些涉及其他表的表达式，使用column_property()，或者当我们不希望默认情况下加载表达式时使用deferred()（很可能就是这种情况）这里）。例如：

from sqlalchemy import *
from sqlalchemy.orm import *
from sqlalchemy.ext.declarative import declarative_base


Base = declarative_base()

class Widget(Base):
    __tablename__ = 'widgets'
    id = Column(Integer, primary_key=True)
    name = Column(String(50))
    sidebar = Column(Integer)

class Dashboard(Base):
    __tablename__ = 'dashboard'
    user_id = Column(Integer, ForeignKey('users.id'), primary_key=True)
    widget_id = Column(Integer, ForeignKey('widgets.id'), primary_key=True)
    delta = Column(Integer)

    widget = relationship('Widget')

    widget_sidebar = deferred(select([Widget.sidebar]).where(Widget.id == widget_id))

class User(Base):
    __tablename__ = 'users'
    id = Column(Integer, primary_key=True)
    login = Column(Unicode(255), unique=True)

    dashboard = relationship('Dashboard', order_by='Dashboard.widget_sidebar, Dashboard.delta')


e = create_engine("sqlite://", echo=True)
Base.metadata.create_all(e)

s = Session(e)

w1, w2 = Widget(name='w1', sidebar=1), Widget(name='w2', sidebar=2)
s.add_all([
    User(login='u1', dashboard=[
        Dashboard(
            delta=1, widget=w1
        ),
        Dashboard(
            delta=2, widget=w2
        )
    ]),
])
s.commit()

print s.query(User).first().dashboard

“.dashboard”加载发出的最终SQL是：

SELECT dashboard.user_id AS dashboard_user_id, dashboard.widget_id AS dashboard_widget_id, dashboard.delta AS dashboard_delta 
FROM dashboard 
WHERE ? = dashboard.user_id ORDER BY (SELECT widgets.sidebar 
FROM widgets 
WHERE widgets.id = dashboard.widget_id), dashboard.delta

请记住，MySQL会为上面的子查询执行糟糕的作业优化。如果您需要高性能，可以考虑将“侧栏”的值复制到“仪表板”中，即使这样会使一致性更难维护。

SQLAlchemy：按关系中的关系字段排序

1 个答案: